Question

If \( f(x) = \int_0^x t(\sin x - \sin t)dt \) then :

• A. \( f'''(x) - f''(x) = \cos x - 2x\sin x \)
• B. \( f'''(x) + f'(x) = \cos x - 2x\sin x \)
• C. \( f'''(x) + f''(x) = \sin x \)
• D. \( f'''(x) + f''(x) - f'(x) = \cos x \)

Hint:

Always extract terms containing 'x' outside the integral before differentiating. If you see a term in the derivative that matches an earlier order derivative, substitute it immediately to find the required differential equation. \

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To differentiate an integral where the variable $x$ appears both in the limits and the integrand, we use the Leibniz Rule. First, we expand the integrand to separate the terms involving $x$.
Step 2: Key Formula or Approach:
Leibniz Rule: \(\frac{d}{dx} \int_{a}^{x} g(x, t) dt = g(x, x) + \int_{a}^{x} \frac{\partial}{\partial x} g(x, t) dt\).
Step 3: Detailed Explanation:
1. Rewrite \(f(x)\): \[ f(x) = \sin x \int_{0}^{x} t \, dt - \int_{0}^{x} t \sin t \, dt \]
2. Differentiate once (\(f'(x)\)): \[ f'(x) = \cos x \int_{0}^{x} t \, dt + \sin x \cdot [x] - [x \sin x] \] \[ f'(x) = \cos x \left[ \frac{x^2}{2} \right] + x \sin x - x \sin x = \frac{x^2}{2} \cos x \]
3. Differentiate again (\(f''(x)\)): \[ f''(x) = \frac{1}{2} [2x \cos x - x^2 \sin x] = x \cos x - \frac{x^2}{2} \sin x \]
4. Differentiate again (\(f'''(x)\)): \[ f'''(x) = [1 \cdot \cos x - x \sin x] - \frac{1}{2} [2x \sin x + x^2 \cos x] \] \[ f'''(x) = \cos x - x \sin x - x \sin x - \frac{x^2}{2} \cos x = \cos x - 2x \sin x - \frac{x^2}{2} \cos x \]
5. Calculate \(f'''(x) + f'(x)\): \[ f'''(x) + f'(x) = (\cos x - 2x \sin x - \frac{x^2}{2} \cos x) + (\frac{x^2}{2} \cos x) = \cos x - 2x \sin x \]
Step 4: Final Answer
The correct relation is \(f'''(x)+f'(x)=\cos x-2x \sin x\).